计算1-1000范围内所有质数的和

A＝5

1、缺一个#include"stdio.h"；2、里面有汉语标点符号,如if(a%i==0）（这个是汉语的）,for(a=2；a

761271000以内质数表23571113171923293137414347535961677173798389971011031071091131271311371391491511571631

PrivateSubCommand1_Click()DimaAsIntegerDimbAsIntegerDimcAsIntegerDimiAsInteger,jAsIntegera=Text1b=Te

PublicSubqqqqq()DimiAsIntegerDimjAsIntegerDimsAsIntegerDimCAsBooleans=0Fori=2To100C=FalseForj=2Toi-1

%用函数>>primes(1000);%用循环R=[];fork=1:1000s=0;forp=1:kifmod(k,p)==0s=s+1;end;end;ifs==2R=[R;k];end;end;

Private Sub Command1_Click()Dim i As Long, x As LongFor 

2357111317192329313741434753596167717379838997101103107109113127131137139149151157163167173179181191

PrivateSubCommand1_Click()Fori=2To1000Forj=2ToiIfiModj=0ThenExitForNextjIfj=iThenPrinti;NextiEndSu

这样就可以了.只要判断第百位跟各位是否一致就行了.publicstaticvoidMain(){intsum=0;for(inti=100;i

一千以内的质数有：2357111317192329313741434753596167717379838997101103107109113127131137139149151157163167173

100以内质数记忆法100以内的质数共有25个,这些质数我们经常用到,可以用下面的两种办法记住它们.一、规律记忆法首先记住2和3,而2和3两个质数的乘积为6.100以内的质数,一般都在6的倍数前、后的

def count_multiples(base, start, stop):    result=[]  &

刚好在做C++,顺便帮你做下,#includemain(){inta,i,sum=0;printf("素数:\n");for(a=1;a

#includevoidmain(){inti,a,s=0;for(i=0;i0){a=k%10;k/=10;sum+=a;summ*=a;}if(sum>su

1、首先你要记住常用的元素化合价（必须记住,没有其它技巧）；如：常见元素的化合价口诀：氢钠钾银+1价,钙钡镁锌+2价（解释：如氢、钠、钾和银均是+1价,其它同理）；一二铜,二三铁,二四六硫二四碳（解释

2357111317192329313741434753596167717379838997

程序计算出来的!classpn{publicstaticvoidmain(String[]args){intcnt=1,pre=2;System.out.println(2);go:for(inti=

2357111317192329313741434753596167717379838997

101,103,107,109,113,127,131,137,139,149,151,157,163,167,173,179,181,191,193,197,199,　　211,223,227,22